# Tagged Questions

For questions about finding factors of e.g. integers or polynomials

2answers
3k views

### Is factoring polynomials as hard as factoring integers?

There seems to be a consensus that factorization of integers is hard (in some precise computational sense.) Is it known whether polynomial factorization is computationally easy or hard? One thing I ...
5answers
3k views

### Proof of $a^n+b^n$ divisible by a+b when n is odd [closed]

I read somewhere that $(a^n - b^n)$ It is always divisible by a-b. When n is even it is also divisible by a+b. When n is odd it is not divisible by a+b. and $(a^n + b^n)$ ...
4answers
542 views

### Factoring $ac$ to factor $ax^2+bx+c$

I was watching a first-year high-school-algebra student struggle with factoring quadratics last night. Given a quadratic $ax^2+bx+c$ (I'll give you the exact example in a moment), her method — ...
4answers
2k views

### How can we prove that among positive integers any number can have only one prime factorization?

I have read right from school that prime factorization is unique, but have never found proof for this. Can someone show me the proof?
12answers
5k views

### Why does factoring eliminate a hole in the limit?

$$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)$$ I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and ...
3answers
1k views

### A theorem about prime divisors of generalized Fermat numbers?

A theorem of Édouard Lucas related to the Fermat numbers states that : Any prime divisor $p$ of $F_n=2^{2^n}+1$ is of the form $p=k\cdot 2^{n+2}+1$ whenever $n$ is greater than one. Does anyone ...
4answers
2k views

### How to factor the quadratic polynomial $2x^2-5xy-y^2$?

How do I factor this polynomial: $2x^2-5xy-y^2$ ?
3answers
455 views

### Multiplication partitioning into k distinct elements

Let's say I have a list with the prime factors of a number $n$ and their corresponding exponents. Is there a formula to calculate how many multiplications with $k$ distinct factors of $n$ are possible?...
2answers
86 views

### Quadratic Polynomial factorization

This could be primary school stuff. But I want to ask it. In factoring $x^2+bx+c$ (i.e. $a = 1$ in $ax^2+bx+c$), we find $m$ and $n$ such that $m+n = b$ and $mn=c$. We can reason this well as follows:...
2answers
1k views

2answers
855 views

### Factoring Quadratics: Asterisk Method

I'm teaching my students about factoring quadratics. We've done GCF, difference of two squares, squared binomials, and grouping. One of my colleagues then found this asterisk method on line. It's ...
8answers
647 views

### How to factor quadratic $ax^2+bx+c$?

How do I shorten this? How do I have to think? $$x^2 + x - 2$$ The answer is $$(x+2)(x-1)$$ I don't know how to get to the answer systematically. Could someone explain? Does anyone have a link to ...
6answers
323 views

### Factoring Quadratics

Is there a method to find which numbers to use when simplifying quadratics? For example $x^2 + 5x + 6$ is easy enough to find, but what if I have bigger numbers, or I have this quadratic expression: ...
2answers
64 views

### If I remove the premise $a\neq b$ in this question, will the statement still be true?

I encountered this proving problem, I can do the proof but my question is why in the condition/premise we need $a$ to be unequal to $b$? My guess is that even $a=b$, the statement is still true, is it ...
16answers
98k views

### What is a real world application of polynomial factoring?

The wife and I are sitting here on a Saturday night doing some algebra homework. We are factoring polynomials and we both had the same thought at the same time: when are we going to use this? I feel ...
5answers
5k views

1answer
81 views

### Probability that a number has $m$ indistinct factors

I just discovered Matlab's factor()-function, and I randomly typed in 20081294819, and to my surprise it only had two factors (5099 and 3938281)! I had expected many more factors for such a big number ...
2answers
64 views

### Finding integers satisfying $m^2 - n^2 = 1111$

We have to find the integers $m$ and $n$ which will satisfy the given condition: $$m^2-n^2=1111.$$ What could be the answer and how? i tried using trial and error and that took a long time.
1answer
186 views

### By viewing the polynomials as a difference of two squares, factorise the following polynomials.

By viewing the polynomials as a difference of two squares, factorise the following polynomial: $$x^4+x^2+1.$$ I searched but couldn't find a way to solve this Edit: By using Hans Lundmark hint, I get:...
1answer
70 views

### Irreducibility of polynomials $x^{2^{n}}+1$

I would like to if the polynomials of the form $x^{2^{n}}+1$ are irreducible over $\mathbb{Q}$ and in that case if there is some "easy" proof for that (where easy means not using a big theory like ...
0answers
19 views

1answer
2k views

### Factorize a Symmetric matrix as an 'Approximation' with an outer product.

(deprecated-taken back based on discussion(OLD)) What is a good way to factor a symmetric matrix $X$ as an outer product of two vectors $u$ and $v$. i.e, Find two vectors $u$ and $v$ such that $X=uv^T$...